Re: ext kalman filter
- To: mathgroup at smc.vnet.net
- Subject: [mg48539] Re: ext kalman filter
- From: jax79sg at msn.com (jax)
- Date: Fri, 4 Jun 2004 04:50:47 -0400 (EDT)
- References: <xpnijmi3ihb7@legacy>, <o4ij0urji8aw@legacy>
- Sender: owner-wri-mathgroup at wolfram.com
On 3 Jun 04 08:29:07 -0400 (EDT), Jonathan Chauvin wrote:
>There can be a lot of possibilities.
>First of all, is the system observable? is it uniformly observable?
>the differential equation has a solution...
>Can you explain us your system too see in more details?
>On 1 Jun 04 04:30:06 -0400 (EDT), putey wrote:
>>Hi, implemented a ekf on my application but within 3 cycles, of
>>aproximation, the values run to infinity. Is this a sign of
>>How do i solve this?
>>Is it related to the follow extract?
>>An important feature of the EKF, and indeed the key to the
>>one-step-at-a-time approach, is that the Jacobian in the equation
>>the Kalman gain serves to correctly propagate or "magnify" only the
>>relevant component of the measurement information. For example, if
>>there is not a one-to-one mapping between the measurement and the
>>state via h(·), the Jacobian affects the Kalman gain so that it
>>magnifies the portion of the residual that does affect the state.
>>course if for all measurements there is not a one-to-one mapping
>>between the measurement and the state via h(·), then as you might
>>expect the filter will quickly diverge. The control theory term to
>>describe this situation is unobservable.
I had similar problems listed here, it seems that one of my inverse
matrix has encountered a zero determinent in the early cycles. My code
did not check for determinent and went ahead with the gaussian
elimination, turn out with numbers near to inf. Does this mean EKF
becomes useless if the matrix to be inverted is uninvertible?
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